How do I parse a string into a number with Dart?

Question

I would like to parse strings like 1 or 32.23 into integers and doubles. How can I do this with Dart?

Answer 1

If the default value is 0 you can use something like double.parse("0${textfieldController.value.text}"); adding a zero before the string you are trying to parse ensuring that no exceptions are raised even when the input is empty or cannot be successfully converted into a numeric value.

Answer 2

for easy reading, you can make it an extension

extension MyCustomString on String {
  double? get strToDouble {
    double? d = double.tryParse(replaceAll(RegExp(r'[^0-9\.]'), ''));
    return d;
  }
  
  int? get strToInt {
    int? i = int.tryParse(replaceAll(RegExp(r'[^0-9]'), ''));
    return i;
  }
}

void main() {
  var myString = '13.2';
  var myDouble = myString.strToDouble;
  print(myDouble);
  print(myDouble.runtimeType);
  
  myString = '13';
  var myInt = myString.strToInt;
  print(myInt);
  print(myInt.runtimeType);
}

https://dartpad.dev/49bde0c1ed780decc902f3d4d06d8f0c?

Answer 3

void main(){
String myString ='111';
int data = int.parse(myString);
print(data);
}

Answer 4

If you don't know whether your type is string or int you can do like this:

int parseInt(dynamic s){
  if(s.runtimeType==String) return int.parse(s);
  return s as int;
}

For double:

double parseDouble(dynamic s){
  if(s.runtimeType==String) return double.parse(s);
  return s as double;
}

Therefore you can do parseInt('1') or parseInt(1)

Answer 5

You can do this for easy conversion like this

Example Code Here

void main() {
  var myInt = int.parse('12345');
 var number = myInt.toInt();
  print(number); // 12345
  print(number.runtimeType);  // int
  
  
    var myDouble = double.parse('123.45');
  var double_int = myDouble.toDouble();
  print(double_int); // 123.45
   print(double_int.runtimeType);  
}

Answer 6

String age = stdin.readLineSync()!; // first take the input from user in string form
int.parse(age);  // then parse it to integer that's it

Answer 7

Above solutions will not work for String like:

String str = '123 km';

So, the answer in a single line, that works in every situation for me will be:

int r = int.tryParse(str.replaceAll(RegExp(r'[^0-9]'), '')) ?? defaultValue;
or
int? r = int.tryParse(str.replaceAll(RegExp(r'[^0-9]'), ''));

But be warned that it will not work for the below kind of string

String problemString = 'I am a fraction 123.45';
String moreProblem = '20 and 30 is friend';

If you want to extract double which will work in every kind then use:

double d = double.tryParse(str.replaceAll(RegExp(r'[^0-9\.]'), '')) ?? defaultValue;
or
double? d = double.tryParse(str.replaceAll(RegExp(r'[^0-9\.]'), ''));

This will work for problemString but not for moreProblem.

Answer 8

Convert String to Int

var myInt = int.parse('12345');
assert(myInt is int);
print(myInt); // 12345
print(myInt.runtimeType);

Convert String to Double

var myDouble = double.parse('123.45');
assert(myInt is double);
print(myDouble); // 123.45
print(myDouble.runtimeType);

Example in DartPad

Answer 9

you can parse string with int.parse('your string value');.

Example:- int num = int.parse('110011'); print(num); // prints 110011 ;

Answer 10

 void main(){
  var x = "4";
  int number = int.parse(x);//STRING to INT

  var y = "4.6";
  double doubleNum = double.parse(y);//STRING to DOUBLE

  var z = 55;
  String myStr = z.toString();//INT to STRING
}

int.parse() and double.parse() can throw an error when it couldn't parse the String

Answer 11

As per dart 2.6

The optional onError parameter of int.parse is deprecated. Therefore, you should use int.tryParse instead.

Note: The same applies to double.parse. Therefore, use double.tryParse instead.

  /**
   * ...
   *
   * The [onError] parameter is deprecated and will be removed.
   * Instead of `int.parse(string, onError: (string) => ...)`,
   * you should use `int.tryParse(string) ?? (...)`.
   *
   * ...
   */
  external static int parse(String source, {int radix, @deprecated int onError(String source)});

The difference is that int.tryParse returns null if the source string is invalid.

  /**
   * Parse [source] as a, possibly signed, integer literal and return its value.
   *
   * Like [parse] except that this function returns `null` where a
   * similar call to [parse] would throw a [FormatException],
   * and the [source] must still not be `null`.
   */
  external static int tryParse(String source, {int radix});

So, in your case it should look like:

// Valid source value
int parsedValue1 = int.tryParse('12345');
print(parsedValue1); // 12345

// Error handling
int parsedValue2 = int.tryParse('');
if (parsedValue2 == null) {
  print(parsedValue2); // null
  //
  // handle the error here ...
  //
}

Answer 12

In Dart 2 int.tryParse is available.

It returns null for invalid inputs instead of throwing. You can use it like this:

int val = int.tryParse(text) ?? defaultValue;

Answer 13

You can parse a string into an integer with int.parse(). For example:

var myInt = int.parse('12345');
assert(myInt is int);
print(myInt); // 12345

Note that int.parse() accepts 0x prefixed strings. Otherwise the input is treated as base-10.

You can parse a string into a double with double.parse(). For example:

var myDouble = double.parse('123.45');
assert(myDouble is double);
print(myDouble); // 123.45

parse() will throw FormatException if it cannot parse the input.